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Mathematics 14 Online
OpenStudy (anonymous):

*Confusion* Rita promises to give her 12-year-old daughter $12,000 when she turns 18. How much does she need to put in the bank now if the interest rate is 12% per year?

jimthompson5910 (jim_thompson5910):

does it mention anything about compounding interest?

OpenStudy (anonymous):

No

jimthompson5910 (jim_thompson5910):

hmm frustrating, it could be compounded monthly or annually (or maybe something else)

OpenStudy (anonymous):

I tried using both future and present value but neither of them gve me an answer i can choose

jimthompson5910 (jim_thompson5910):

oh you're given a list of choices? maybe that will help me figure out what they want

jimthompson5910 (jim_thompson5910):

sometimes they think students are mindreaders which is frustrating

OpenStudy (anonymous):

6079.58 5248.19 6502.12 is what they give me

OpenStudy (anonymous):

its very frustrating

jimthompson5910 (jim_thompson5910):

ok so through trial and error, I figured out they are using compound interest with the interest being compounded 1 time per year

jimthompson5910 (jim_thompson5910):

do you know the compound interest formula?

OpenStudy (anonymous):

that may be why, I'm not familiar with it lol

jimthompson5910 (jim_thompson5910):

the formula is \[\Large A = P(1+r/n)^{nt}\] In this case, A = 12000 P = unknown r = 0.12 n = 1 (compounding 1 time per year) t = 6 years (difference from 12 to 18)

OpenStudy (anonymous):

i will have to note this

jimthompson5910 (jim_thompson5910):

so plugging all that in gives \[\Large A = P(1+r/n)^{nt}\] \[\Large 12000 = P(1+0.12/1)^{1*6}\] \[\Large 12000 = P(1+0.12)^{6}\] \[\Large 12000 = P(1.12)^{6}\] You keep going to solve for P

OpenStudy (anonymous):

thank you thank you thank you so very much :)

jimthompson5910 (jim_thompson5910):

you're welcome

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